I know dimension is the number of vectors in the basis, so I want to know if W's vectors are each linearly independent...

But I have no convenient way of telling if the vectors are linearly independent. The Wronskian is no good - it just evaluates to 0 for various substitutions ($x=0,1)$, which doesn't give me information. I'm looking for an efficient way to answer this problem.

I've also considered the rank-nullity theorem. dim$(P_2) = 3$, but does that tell me anything about the dim$(W)?$

2 Answers
2

Knowing that the dimension of $P_2$ is $3$ lets you know that the dimension of $W$ is at most $3$. Pick any $2$ of the vectors you're given and observe that they're linearly independent, so that the dimension of $W$ is at least $2$.

To determine whether they're linearly independent, observe the following. Let's say $p_1=2+x^2,$ $p_2=4-2x+3x^2,$ $p_3=1+x.$ Then we can get rid of all $x^2$ terms by taking $p_2-3p_1=-2-2x=-2p_3$. Thus, they aren't linearly independent, so $W$ can't have dimension $3,$ and so must have dimension $2$.

More generally (and more easily generalized), let's suppose that $a_1,a_2,a_3$ are constants such that $$0=a_1p_1+a_2p_2+a_3p_3=(2a_1+4a_2+a_3)+(-2a_2+a_3)x+(a_1+3a_2)x^2,$$ which means that each of $2a_1+4a_2+a_3,$ $-2a_2+a_3,$ and $a_1+3a_2$ is zero. Put into matrix form (with $1,x,x^2$ as our standard ordered basis), $$\left[\begin{array}{ccc}2 & 4 & 1\\0 & -2 & 1\\1 & 3 & 0\end{array}\right]\left[\begin{array}{c}a_1\\a_2\\a_3\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right].$$ Thus, the problem reduces to determining the rank of the matrix. (I leave it to you to determine that it's $2$.)

May be you can try this. If $v_1,v_2,v_3$ are your given vectors, they are linearly independent iff there exist $c_1,c_2,c_3$ such that the only solution to $c_1v_1+c_2v_2+c_3v_3=0$ is $c_i=0,\forall i$.

Using this, your set of polynomials can be combined to give
\begin{align}
(c_1+3c_2)x^2+(-2c_2+c_3)x+(2c_1+4c_2+c_3)=0
\end{align}